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--- lecture_16_slides [2016/03/30 15:37] – [Exercise] rupert
+++ lecture_16_slides [2017/04/03 15:26] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
-==== Example: $n=3$ ====
-Let $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$.
-  * Matrix of signs: $\mat{+&-&+\\-&+&-\\+&-&+}$
-  * Matrix of cofactors: $C=\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$
-  * So the adjoint of $A$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$.
-==== ====
+===== Finding the inverse of an invertible $n\times n$ matrix =====
-  * Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
-  * $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
+==== The adjoint of a square matrix ====
-  * $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
-  * So $A^{-1}=-J$.
+{{page>adjoint}}
-  * And $\det(A)=-1$.
-  * So $A^{-1}=\frac1{\det(A)}J$ again.
 ==== Theorem: key property of the adjoint of a square matrix ====
@@ Line 31: / Line 23: @@
   * Divide the equation $AJ=(\det A)I_n=JA$ by $\det A$.
-  * $A(\frac1{\det A}J)=I_n=(\frac1{\det A})JA$
+  * $A(\frac1{\det A}J)=I_n=(\frac1{\det A}J)A$
   * So $A^{-1}=\frac1{\det A} J$.  ■
+==== Example: $n=2$ ====
+If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{a&b\\c&d}$, then $C=\mat{d&-c\\-b&a}$, so the adjoint of $A$ is $J=C^T=\mat{d&-b\\-c&a}$.
+  * So if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J=\frac1{ad-bc}\mat{d&-b\\-c&a}$.
+  * Same as previous formula!
+==== Example: $n=3$ ====
+Find $J$, the adjoint of $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, and compute $AJ$, $JA$, $\det(A)$ and $A^{-1}$.
+  * Matrix of signs: $\mat{+&-&+\\-&+&-\\+&-&+}$
+  * Matrix of cofactors: $C=\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$
+==== ====
+  * Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
+  * $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
+  * $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
+  * So $\det(A)=-1$.
+  * So $A^{-1}=\frac1{\det(A)}J=-J=\mat{4&-11&-12\\-2&6&7\\-3&9&10}$.
 ==== Example ($n=4$) ====
@@ Line 47: / Line 59: @@
   * So $A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}$.
   * (Easy to check that $AA^{-1}=I_4=A^{-1}A$.)
 ===== A more efficient way to find $A^{-1}$ =====
@@ Line 101: / Line 110: @@
   * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.
-====== Chapter 3: Vectors and geometry ======
-==== Vectors ====
-  * $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a $2\times 1$ column vector
-  * i.e., a pair of numbers written in a column
-  * We also use pairs of numbers to write points in the plane $\mathbb R^2$
-  * e.g., $(4,3)$ is a point
-    * you get there by starting from the origin, moving $4$ units to the right and $3$ units up.
-  * We think $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up.
-  * This movement is called "translation by $\vec v$".
-==== Translation by $\vec v$ ====
-The vector $\vec v=\m{4\\3}$ moves:
-  * $(0,0)$ to $(4,3)$
-  * $(-2,6)$ to $(2,9)$
-  * $(x,y)$ to $(x+4,y+3)$.
-  * We're not too fussy about the difference between points like $(4,3)$ and vectors like $\m{4\\3}$.
-  * If we write points as column vectors, we can perform algebra (addition, subtraction, scalar multiplication) using points and column vectors.
-==== ====
-For example, we could rewrite the examples above by saying that $\vec v=\m{4\\3}$ moves:
-  * $\m{0\\0}$ to $\m{0\\0}+\m{4\\3}=\m{4\\3}$
-  * $\m{-2\\6}$ to $\m{-2\\6}+\m{4\\3}=\m{2\\9}$
-  * $\m{x\\y}$ to $\m{x\\y}+\m{4\\3}=\m{x+4\\y+3}$.
-  * More generally: a column vector $\vec v$ moves a point $\vec x$ to $\vec x+\vec v$.
-==== Example ====
-Which vector $\vec v$ moves the point $A=(-1,3)$ to $B=(5,-4)$?
-  * We need a vector $\vec v$ with $A+\vec v=B$
-  * So $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$.
-  * We write $\vec{AB}=\m{6\\-7}$, since this is the vector which moves $A$ to $B$.
-==== Definition of $\vec{AB}$ ====
-If $A$ and $B$ are any points in $\mathbb{R}^n$, then the vector $\vec{AB}$ is defined by
-\[ \vec{AB}=B-A\]
-(on the right, we interpret points as column vectors so we can subtract them to get a column vector).
-  * $\vec{AB}$ is the vector which moves $A$ to $B$.
-==== Example ====
-In $\mathbb{R}^3$,
-  * if $A=(3,-4,5)$
-  * and $B=(11,6,-2)$
-  * then $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.
-==== The uses of vectors ====
-Vectors are used in geometry and science to represent quantities with both a **magnitude** (size/length) and a **direction**. For example:
-  * displacements (in geometry)
-  * velocities
-  * forces
-Recall that a column vector moves points. Its magnitude, or length, is how far it moves points.
-==== Definition: the length of a vector ====
-If $\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ is a column vector in $\mathbb{R}^n$, then its **magnitude**, or **length**, or **norm**, is the number
-\[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\]
-==== Examples ====
-  * $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
-  * \begin{align*}\left\|\m{1\\0\\-2\\3}\right\|&=\sqrt{1^2+0^2+(-2)^2+3^2}\\&=\sqrt{1+0+4+9}=\sqrt{14}\end{align*}
-==== Exercise ====
-Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then
-\[ \|c\vec v\|=|c|\,\|\vec v\|.\]
-That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.
-  * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, and $\sqrt{c^2}=|c|$.
-==== Distance between two points ====
-$\|\vec{AB}\|$ is the distance from point $A$ to point $B$
-  * since this is the length of vector which takes point $A$ to point $B$.
-  * e.g. how far from from $A=(1,2)$ to $B=(-3,4)$?
-  * $\small\|\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}$.
-  * e.g. what's the length of the main diagonal of the unit cube in $\mathbb{R}^3$?
-  * = distance from $0=(0,0,0)$ to $A=(1,1,1)$
-  * $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.